Question

among 51 - to 56 - year - olds, 29% say they have written an editorial letter while under the influence of alcohol. suppose seven 51 - to 56 - year - olds are selected at random. complete parts (a) through (d) below. (a) what is the probability that all seven have written an editorial letter while under the influence of alcohol? 0.0002 (round to four decimal places as needed.) (b) what is the probability that at least one has not written an editorial letter while under the influence of alcohol? 0.9998 (round to four decimal places as needed.) (c) what is the probability that none of the seven have written an editorial letter while under the influence of alcohol? (round to four decimal places as needed.)

Explanation:

Step1: Identify success - failure probabilities

The probability that a 51 - to 56 - year - old has written an editorial letter while under the influence of alcohol is $p = 0.29$, so the probability that a 51 - to 56 - year - old has not written such a letter is $q=1 - p=1 - 0.29 = 0.71$.

Step2: Calculate the probability that none have written

We use the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success, and $q$ is the probability of failure. Here, $n = 7$, $k = 0$, $p = 0.29$, and $q = 0.71$. Since $C(n,0)=1$, the probability that none of the seven have written an editorial letter while under the influence of alcohol is $P(X = 0)=q^{n}=(0.71)^{7}$.
$(0.71)^{7}=0.0886$.

Answer:

$0.0886$